The duration of shoppers' time in Browse Wrld's new retail outlets is normally distributed with a mean of 27.8 minutes and a standard deviation of 11.4 minutes. How long must a visit be to put a shopper in the longest 10 percent

A visit must be of at least 42.39 minutes to put a shopper in the longest 10 percent.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 27.8 minutes and a standard deviation of 11.4 minutes.

This means that

How long must a visit be to put a shopper in the longest 10 percent?

The 100 - 10 = 90th percentile, which is X when Z has a p-value of 0.9, so X when Z = 1.28.

A visit must be of at least 42.39 minutes to put a shopper in the longest 10 percent.

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step-by-step explanation:

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